How do I find the area between 3 curves?

Question

I have three equations: $y=3/x$, $y=12x$, and $y=x/12$, $x>0$. I am not sure how to go about integrating an equation once I find the intersections. Do I need multiple integrals?

Answer 1

Yes. After drawing a diagram, you should be able to see that the desired area is: $$ \int_0^{1/2} (12x - \tfrac{x}{12}) \, dx + \int_{1/2}^6 (\tfrac{3}{x} - \tfrac{x}{12}) \, dx $$

Answer 2

When you draw a graph of the three curves you notice the hyperbola is cut by 2 radial rays, so better to adopt polar coordinates.

Let $$ \tan^{-1}(1 /12) = t1 ,\tan^{-1}(12) = t2 $$

$$ x \, y = 3 \rightarrow r^2 \sin ( 2 t) = 6 ,\, r^2/2 = 3/ \sin ( 2 t) $$

$$ Area = \int_{t_1}^{t_2} r^2/2 \, dt $$

Can you proceed with it further?